Question

Suppose you stand with one foot on ceramic flooring and one foot on a wool carpet, making contact over an area of 80.0 cm2 with each foot. Both the ceramic and the carpet are 2.00 cm thick and are 10.0∘C on their bottom sides. At what rate must heat transfer occur from each foot to keep the top of the ceramic and carpet at 33.0∘C?

Answer 1

To solve this problem it is necessary to apply the concepts related to the rate of heat transfer in the materials.

By definition the rate of heat conduction transferred is understood as

$\frac{Q}{t} = \frac{kA(T_2-T_1)}{d}$

Where,

$\frac{Q}{t} =$ Rate of heat transfer

k = Thermal conductivity

A = Surface area

d = Thickness

$T_{1,2} =$ Temperature at different states.

Our values are given as,

$k_w = 0.04J/s.m.\°C\\k_c = 0.84J/s.m.\°C\\A = 80*10^{-4}m^2\\d= 2cm\\T_2 = 33\°CT_1 = 10\°C$

Substituting for the wood carpet we have,

$\frac{Q}{t} = \frac{kA(T_2-T_1)}{d}\\\frac{Q}{t} = \frac{(0.04)(80*10^{-4})(33-10)}{2*10^{-2}}\\\frac{Q}{t} = 0.369W$

Substituting for the ceramic we have,

$\frac{Q}{t} = \frac{kA(T_2-T_1)}{d}\\\frac{Q}{t} = \frac{(0.84)(80*10^{-4})(33-10)}{2*10^{-2}}\\\frac{Q}{t} = 7.728W$

Therefore the rate of heat transfer for the wood carpet is 0.37W while the rate of heat transfer for the ceramic is 7.7W